The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives $22.2$ years; the standard deviation is $4.5$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a tiger living less than $17.7$ years.
Explanation: $22.2$ $17.7$ $26.7$ $13.2$ $31.2$ $8.7$ $35.7$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $22.2$ years. We know the standard deviation is $4.5$ years, so one standard deviation below the mean is $17.7$ years and one standard deviation above the mean is $26.7$ years. Two standard deviations below the mean is $13.2$ years and two standard deviations above the mean is $31.2$ years. Three standard deviations below the mean is $8.7$ years and three standard deviations above the mean is $35.7$ years. We are interested in the probability of a tiger living less than $17.7$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the tigers will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the tigers will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $17.7$ years and the other half $({16\%})$ will live longer than $26.7$ years. The probability of a particular tiger living less than $17.7$ years is ${16\%}$.